How do you find the vertex and intercepts for #y=2x^2-16x+27#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer EZ as pi May 7, 2016 #y = 2(x - 4)^2 - 5" "# Vertex is at #(4, -5) # #y-#intercept is at 27 Explanation: #y = 2x^2 - 16x + 27# #y = 2(x^2 - 8x + 27/2) " "#now complete the square #y = 2[color(blue)(x^2 - 8x + 16) -16 +27/2] " " rArr# +16-16 = 0 #y = 2[color(blue)((x - 4)^2) -16 +27/2] " "# #-16+27/2 = -5/2# #y = 2(x-4)^2 - 2xx5/2# #y = 2(x - 4)^2 - 5# Answer link Related questions What is the Vertex Form of a Quadratic Equation? How do you find the vertex form of a quadratic equation? How do you graph quadratic equations written in vertex form? How do you write #y+1=-2x^2-x# in the vertex form? How do you write the quadratic equation given #a=-2# and the vertex #(-5, 0)#? What is the quadratic equation containing (5, 2) and vertex (1, –2)? How do you find the vertex, x-intercept, y-intercept, and graph the equation #y=-4x^2+20x-24#? How do you write #y=9x^2+3x-10# in vertex form? What is the vertex of #y=-1/2(x-4)^2-7#? What is the vertex form of #y=x^2-6x+6#? See all questions in Vertex Form of a Quadratic Equation Impact of this question 1156 views around the world You can reuse this answer Creative Commons License